Question

If 75 moles of BaO are used, how many grams of H2SO4 must be used in the process?

Answer 1

can you write down the reaction between these two?

Answer 2

First off, you need to write the equation. BaO is a base, and H2SO4 is an acid.
Base + Acid = Salt + H2O (water)
To know what salt will be made, in these equations the metal ion (Ba) replaces the hydrogen ion ('H' in H2SO4)

So the equation will be:
BaO + H2SO4 = BaSO4 + H2O

It is already balanced so we will not need to balance this equation, however you always need to check the equation and make sure it is balanced before you continue.

75 moles is a massive amount, but if that's what your question says then we will carry on. If not then just correct yourself and plug in where ever it says '75' in my next calculations with the actual no. of moles. :)

First off we need to find the moles of the desired compound, which is H2SO4.

So, as moles cant be created, the moles of each compound in the equation will be the same (this applies to all equations).
This means that because there is ONE lot of every compound in this equation (there are no numbers in front of any of them, meaning there is 1 of each), the moles of H2SO4 will be 75mol, same as BaO.

If there was a two in front of H2SO4 and a one in front of BaO, we would have to multiply the moles of BaO by two as there are TWO lots of H2SO4, and only ONE lot of BaO

So moles of H2SO4 = 75mol

To work out mass, you need to use this equation:
mass = moles x molar mass

Firstly work out the molar mass of H2SO4:
(1x2) + 32.1 + (16 x 4) = 98.1g mol^-1

Now you can work out the mass:
75mol x 98.1g mol^-1 = 7357.5g

So 7357.5g of H2SO4 is needed (if 75 mol is correct, as that mass is pretty big)

I hope this helps :)